d, ĐK:\(n+2\ne0\Leftrightarrow n\ne-2\)
\(e,A=2\\ \Leftrightarrow\dfrac{n+9}{n+2}=2\\ \Rightarrow n+9=2n+4\\ \Leftrightarrow n=5\\ A=4\\ \Leftrightarrow\dfrac{n+9}{n+2}=4\\ \Leftrightarrow n+9=4n+8\\ \Leftrightarrow3n=1\\ \Leftrightarrow n=\dfrac{1}{3}\)
\(f,A\in Z\\ \Rightarrow\dfrac{n+9}{n+2}\in Z\\ \Rightarrow\dfrac{n+2+7}{n+2}\in Z\\ \Rightarrow1+\dfrac{7}{n+2}\in Z\)
Để \(A\in Z\Rightarrow\dfrac{7}{n+2}\in Z\Rightarrow7⋮\left(n+2\right)\Rightarrow n+2\inƯ\left(7\right)\)
Ta có bảng:
n+2 | -7 | -1 | 1 | 7 |
n | -9 | -3 | -1 | 5 |
Vậy \(n\in\left\{-9;-3;-1;5\right\}\)